Finding Max and Mins Part V

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Hckyplayer8

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Find the critical numbers of f(x) = x1/3 - x 1/5 and determine the absolute max and min values on the interval [0,2] and where they occur.

For f'(x) I have 1/3x2/3 - 1/5x4/5

f(0) = 0
f(2) = 21/3 - 21/5 = ?

That is about as far as I can get in this problem.
 
Hckyplayer8 said:
Find the critical numbers of f(x) = x1/3 - x 1/5 and determine the absolute max and min values on the interval [0,2] and where they occur.

For f'(x) I have 1/3x2/3 - 1/5x4/5

f(0) = 0
f(2) = 21/3 - 21/5 = ?............................ you'll need to use a calculator for this part

That is about as far as I can get in this problem.
Click to expand...
Your expression for f'(x) is incorrect. That should be:

f'(x) = (1/3)x-2/3 - (1/5)x-4/5 ......................... the exponents are negative.

f'(x) = (1/15) * x-4/5 * [5 * x2/15 - 3] .............. continue
 
the derivative of x^n is n*x^(n-1)
which in this case would be
x^1/3 ---> (1/3)*x^(1/3-1)=1/3*x^-2/3

now 1/3x-2/3 - 1/5x-4/5 =0
5x-2/3=3x-4/5
5/x2/3 =3/x4/5

here you see if I go toward zero I will have a problem since the derivative would be going toward infinity. but if you go from either side of zero toward zero (negative or positive) the sign would be positive so there is no discontinuity in our function
we can continue:
x4/5/x2/3=3/5 or x(4/5-2/3)=3/5 ---> x=(3/5)15/2
 
If you subtract 1 from a number the result WILL NOT GET LARGER! You need to catch those mistakes!
 
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